1

physics course

Dr Zyad Ahmed Tawfik

Email zmohammedinayaedusa

Website zyadinayawordpresscom

2

Course Description This course serves as an introduction to the basic principles of physics and also this

course is designed for students in Health Science to enable them to appreciate the

basic concepts of Physics which are relevant to their further studies

Student Learning Outcomes

Upon completion of this course the students are expected to

1 understand the essential elements of physics needed by premedical students

2 Recognize the basic principles of physics in the branches of mechanics movement forces fluid mechanics electric and magnetic phenomena and

radiation 3 Describe the nature phenomena by using the language of physics

4 develop the ability to solve problems and think critically by applying the acquired knowledge of physics to the various problems

5 know how to conduct a series of practical experiments for the study of

physical phenomena related to some previous knowledge

3

Week 2 Daily Objectives SLO

Assignments amp Activities

Week 2

Newtonrsquos laws of Motion

Definition of force Equilibrium state Newtonrsquos laws of motion

Fundamental forces Weight

and friction

11

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 3 Daily Objectives SLO

Assignments amp Activities

Week 3

Review and resolve quizzes on vector

and Newtons laws

11

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 4 Daily Objectives SLO

Assignments amp Activities

Week 4

Work Energy and Power

Work

Kinetic energy

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 5 Daily Objectives SLO

Assignments amp Activities

Weekly Outline of Curriculum

Week 1 Daily Objectives SLO

Assignments amp Activities

Week 1

Vectors

-Addition Geometrical method amp

Analytical method

-Product of vector Scalar and Cross

product

11

- Lecturing - Team work - Exercises - Self-learning - Group Discussions

4

Week 5

Work Energy and Power

Conservative forces and potential energy

Observations of work and energy

Power

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 6 Daily Objectives SLO Assignments amp Activities

Week 6

Work Energy and Power

Review and resolve quizzes on work

and energy

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 7 Daily Objectives SLO Assignments amp Activities

Week 7

Exam1_Examination 11

12

Comprehensive Examination

Good luck

5

Week 8 Daily Objectives SLO Assignments amp Activities

Week 8

Mechanics of non-Viscous Fluid

The Equation of continuity Stream line flow

Bernoullirsquos equation and its static consequences

13

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 9 Daily Objectives SLO Assignments amp Activities

Week 9

Mechanics of non-Viscous Fluid

Role of gravity in blood circulation

Pressure measurement using manometer

13 - Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 10 Daily Objectives SLO Assignments amp Activities

Week 10

Direct Electric Current (DC)

Basic concept of DC Electric current

Ohmrsquos law Electric safety

14

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 11 Daily Objectives SLO Assignments amp Activities

Week 11

Nerve Conduction

Structure of nerve

Electric characteristics of

axon

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 12 Daily Objectives SLO Assignments amp Activities

Week 12

Exam2_Examination

13

14

15

Comprehensive Examination

Good luck

Week 13 Daily Objectives SLO Assignments amp Activities

6

Week 13

Nerve Conduction

Ionic concentration and the resting potential

Response to weak stimuli The action potential

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 14 Daily Objectives SLO

Assignments amp Activities

Week 14

Wave properties of light

The Index of Refraction Reflection of Light Refraction of Light

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 15 Daily Objectives SLO

Assignments amp Activities

Week 15

Wave properties of light

Ionizing Radiation

The Interaction of Radiation with Matter

Chronic Radiation Exposure

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 16 Daily Objectives SLO

Assignments amp Activities

Week 16

Final Examination-Practical Lab

Week 17 Daily Objectives SLO

Assignments amp Activities

Week 17

Final Examination-

Comprehensive

11

12

13

14

15

16

22

23

24

31

32

Comprehensive Final Examination

Must be passed with a score of 60 or

better to pass course

7

Vocabulary

ENGLISH ARABIC

Acceleration تسارع ndashعجلة

Activity نشاط أشعاعى

Air pressure ضغط الهواء

Ampere أمبير

Analytical تحليلى

Analytical Method الطريق التحليلية

Angle زاوية

Angle of Deviation زاوية االنحراف

Angle of Incidence زواية السقوط

Angle of reflection زاوية االنعكاس

Angle of refraction زاوية األنكسار

Archimedess Principle مبدأ أرخميدس

Atmospheric pressure الضغط الجوي

Atom الذرة

Axoplasm جبلة المحوار

Axons محاور عصبية

Balanced Force القوة المتزنة

Bernoullis Principle law مبدأ )قانون( برنولي

Binding Energy طاقة الربط

Blood pressure ضغط الدم

blood vessel وعاء دموى

Buoyancy الطفو

Buoyant force قوة الطفو

8

Capacitance سعة المكثف

Capacitor المكثف

Coefficient of friction معامل األحتكاك

Charge شحنة

Circuit دائرة

cohesion قوة التماسك بين جزئيات السائل

components of a vector مكونات متجه

compression ضغط كباس أنضغاط

conduction توصيل

conductivity معامل الموصلية الحرارية

Conductor الموصل

Conservation of Energy حفظ الطاقة

conservation of momentum حفظ كمية الحركة

consumed مستهلك

Coulomb كولوم

cube مكعب

Current تيار

deceleration تباطؤ

Density الكثافة

Dendrites التشعبات العصبية

diffraction حيود الضوء

dimensions االبعاد

Dispersion تشتت تفرق

direction إتجاه

directly proportional يتناسب طرديا

displacement اإلزاحة

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

3

Week 2 Daily Objectives SLO

Assignments amp Activities

Week 2

Newtonrsquos laws of Motion

Definition of force Equilibrium state Newtonrsquos laws of motion

Fundamental forces Weight

and friction

11

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 3 Daily Objectives SLO

Assignments amp Activities

Week 3

Review and resolve quizzes on vector

and Newtons laws

11

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 4 Daily Objectives SLO

Assignments amp Activities

Week 4

Work Energy and Power

Work

Kinetic energy

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 5 Daily Objectives SLO

Assignments amp Activities

Weekly Outline of Curriculum

Week 1 Daily Objectives SLO

Assignments amp Activities

Week 1

Vectors

-Addition Geometrical method amp

Analytical method

-Product of vector Scalar and Cross

product

11

- Lecturing - Team work - Exercises - Self-learning - Group Discussions

4

Week 5

Work Energy and Power

Conservative forces and potential energy

Observations of work and energy

Power

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 6 Daily Objectives SLO Assignments amp Activities

Week 6

Work Energy and Power

Review and resolve quizzes on work

and energy

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 7 Daily Objectives SLO Assignments amp Activities

Week 7

Exam1_Examination 11

12

Comprehensive Examination

Good luck

5

Week 8 Daily Objectives SLO Assignments amp Activities

Week 8

Mechanics of non-Viscous Fluid

The Equation of continuity Stream line flow

Bernoullirsquos equation and its static consequences

13

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 9 Daily Objectives SLO Assignments amp Activities

Week 9

Mechanics of non-Viscous Fluid

Role of gravity in blood circulation

Pressure measurement using manometer

13 - Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 10 Daily Objectives SLO Assignments amp Activities

Week 10

Direct Electric Current (DC)

Basic concept of DC Electric current

Ohmrsquos law Electric safety

14

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 11 Daily Objectives SLO Assignments amp Activities

Week 11

Nerve Conduction

Structure of nerve

Electric characteristics of

axon

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 12 Daily Objectives SLO Assignments amp Activities

Week 12

Exam2_Examination

13

14

15

Comprehensive Examination

Good luck

Week 13 Daily Objectives SLO Assignments amp Activities

6

Week 13

Nerve Conduction

Ionic concentration and the resting potential

Response to weak stimuli The action potential

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 14 Daily Objectives SLO

Assignments amp Activities

Week 14

Wave properties of light

The Index of Refraction Reflection of Light Refraction of Light

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 15 Daily Objectives SLO

Assignments amp Activities

Week 15

Wave properties of light

Ionizing Radiation

The Interaction of Radiation with Matter

Chronic Radiation Exposure

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 16 Daily Objectives SLO

Assignments amp Activities

Week 16

Final Examination-Practical Lab

Week 17 Daily Objectives SLO

Assignments amp Activities

Week 17

Final Examination-

Comprehensive

11

12

13

14

15

16

22

23

24

31

32

Comprehensive Final Examination

Must be passed with a score of 60 or

better to pass course

7

Vocabulary

ENGLISH ARABIC

Acceleration تسارع ndashعجلة

Activity نشاط أشعاعى

Air pressure ضغط الهواء

Ampere أمبير

Analytical تحليلى

Analytical Method الطريق التحليلية

Angle زاوية

Angle of Deviation زاوية االنحراف

Angle of Incidence زواية السقوط

Angle of reflection زاوية االنعكاس

Angle of refraction زاوية األنكسار

Archimedess Principle مبدأ أرخميدس

Atmospheric pressure الضغط الجوي

Atom الذرة

Axoplasm جبلة المحوار

Axons محاور عصبية

Balanced Force القوة المتزنة

Bernoullis Principle law مبدأ )قانون( برنولي

Binding Energy طاقة الربط

Blood pressure ضغط الدم

blood vessel وعاء دموى

Buoyancy الطفو

Buoyant force قوة الطفو

8

Capacitance سعة المكثف

Capacitor المكثف

Coefficient of friction معامل األحتكاك

Charge شحنة

Circuit دائرة

cohesion قوة التماسك بين جزئيات السائل

components of a vector مكونات متجه

compression ضغط كباس أنضغاط

conduction توصيل

conductivity معامل الموصلية الحرارية

Conductor الموصل

Conservation of Energy حفظ الطاقة

conservation of momentum حفظ كمية الحركة

consumed مستهلك

Coulomb كولوم

cube مكعب

Current تيار

deceleration تباطؤ

Density الكثافة

Dendrites التشعبات العصبية

diffraction حيود الضوء

dimensions االبعاد

Dispersion تشتت تفرق

direction إتجاه

directly proportional يتناسب طرديا

displacement اإلزاحة

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

4

Week 5

Work Energy and Power

Conservative forces and potential energy

Observations of work and energy

Power

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 6 Daily Objectives SLO Assignments amp Activities

Week 6

Work Energy and Power

Review and resolve quizzes on work

and energy

12

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 7 Daily Objectives SLO Assignments amp Activities

Week 7

Exam1_Examination 11

12

Comprehensive Examination

Good luck

5

Week 8 Daily Objectives SLO Assignments amp Activities

Week 8

Mechanics of non-Viscous Fluid

The Equation of continuity Stream line flow

Bernoullirsquos equation and its static consequences

13

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 9 Daily Objectives SLO Assignments amp Activities

Week 9

Mechanics of non-Viscous Fluid

Role of gravity in blood circulation

Pressure measurement using manometer

13 - Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 10 Daily Objectives SLO Assignments amp Activities

Week 10

Direct Electric Current (DC)

Basic concept of DC Electric current

Ohmrsquos law Electric safety

14

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 11 Daily Objectives SLO Assignments amp Activities

Week 11

Nerve Conduction

Structure of nerve

Electric characteristics of

axon

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 12 Daily Objectives SLO Assignments amp Activities

Week 12

Exam2_Examination

13

14

15

Comprehensive Examination

Good luck

Week 13 Daily Objectives SLO Assignments amp Activities

6

Week 13

Nerve Conduction

Ionic concentration and the resting potential

Response to weak stimuli The action potential

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 14 Daily Objectives SLO

Assignments amp Activities

Week 14

Wave properties of light

The Index of Refraction Reflection of Light Refraction of Light

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 15 Daily Objectives SLO

Assignments amp Activities

Week 15

Wave properties of light

Ionizing Radiation

The Interaction of Radiation with Matter

Chronic Radiation Exposure

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 16 Daily Objectives SLO

Assignments amp Activities

Week 16

Final Examination-Practical Lab

Week 17 Daily Objectives SLO

Assignments amp Activities

Week 17

Final Examination-

Comprehensive

11

12

13

14

15

16

22

23

24

31

32

Comprehensive Final Examination

Must be passed with a score of 60 or

better to pass course

7

Vocabulary

ENGLISH ARABIC

Acceleration تسارع ndashعجلة

Activity نشاط أشعاعى

Air pressure ضغط الهواء

Ampere أمبير

Analytical تحليلى

Analytical Method الطريق التحليلية

Angle زاوية

Angle of Deviation زاوية االنحراف

Angle of Incidence زواية السقوط

Angle of reflection زاوية االنعكاس

Angle of refraction زاوية األنكسار

Archimedess Principle مبدأ أرخميدس

Atmospheric pressure الضغط الجوي

Atom الذرة

Axoplasm جبلة المحوار

Axons محاور عصبية

Balanced Force القوة المتزنة

Bernoullis Principle law مبدأ )قانون( برنولي

Binding Energy طاقة الربط

Blood pressure ضغط الدم

blood vessel وعاء دموى

Buoyancy الطفو

Buoyant force قوة الطفو

8

Capacitance سعة المكثف

Capacitor المكثف

Coefficient of friction معامل األحتكاك

Charge شحنة

Circuit دائرة

cohesion قوة التماسك بين جزئيات السائل

components of a vector مكونات متجه

compression ضغط كباس أنضغاط

conduction توصيل

conductivity معامل الموصلية الحرارية

Conductor الموصل

Conservation of Energy حفظ الطاقة

conservation of momentum حفظ كمية الحركة

consumed مستهلك

Coulomb كولوم

cube مكعب

Current تيار

deceleration تباطؤ

Density الكثافة

Dendrites التشعبات العصبية

diffraction حيود الضوء

dimensions االبعاد

Dispersion تشتت تفرق

direction إتجاه

directly proportional يتناسب طرديا

displacement اإلزاحة

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

5

Week 8 Daily Objectives SLO Assignments amp Activities

Week 8

Mechanics of non-Viscous Fluid

The Equation of continuity Stream line flow

Bernoullirsquos equation and its static consequences

13

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 9 Daily Objectives SLO Assignments amp Activities

Week 9

Mechanics of non-Viscous Fluid

Role of gravity in blood circulation

Pressure measurement using manometer

13 - Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 10 Daily Objectives SLO Assignments amp Activities

Week 10

Direct Electric Current (DC)

Basic concept of DC Electric current

Ohmrsquos law Electric safety

14

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 11 Daily Objectives SLO Assignments amp Activities

Week 11

Nerve Conduction

Structure of nerve

Electric characteristics of

axon

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 12 Daily Objectives SLO Assignments amp Activities

Week 12

Exam2_Examination

13

14

15

Comprehensive Examination

Good luck

Week 13 Daily Objectives SLO Assignments amp Activities

6

Week 13

Nerve Conduction

Ionic concentration and the resting potential

Response to weak stimuli The action potential

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 14 Daily Objectives SLO

Assignments amp Activities

Week 14

Wave properties of light

The Index of Refraction Reflection of Light Refraction of Light

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 15 Daily Objectives SLO

Assignments amp Activities

Week 15

Wave properties of light

Ionizing Radiation

The Interaction of Radiation with Matter

Chronic Radiation Exposure

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 16 Daily Objectives SLO

Assignments amp Activities

Week 16

Final Examination-Practical Lab

Week 17 Daily Objectives SLO

Assignments amp Activities

Week 17

Final Examination-

Comprehensive

11

12

13

14

15

16

22

23

24

31

32

Comprehensive Final Examination

Must be passed with a score of 60 or

better to pass course

7

Vocabulary

ENGLISH ARABIC

Acceleration تسارع ndashعجلة

Activity نشاط أشعاعى

Air pressure ضغط الهواء

Ampere أمبير

Analytical تحليلى

Analytical Method الطريق التحليلية

Angle زاوية

Angle of Deviation زاوية االنحراف

Angle of Incidence زواية السقوط

Angle of reflection زاوية االنعكاس

Angle of refraction زاوية األنكسار

Archimedess Principle مبدأ أرخميدس

Atmospheric pressure الضغط الجوي

Atom الذرة

Axoplasm جبلة المحوار

Axons محاور عصبية

Balanced Force القوة المتزنة

Bernoullis Principle law مبدأ )قانون( برنولي

Binding Energy طاقة الربط

Blood pressure ضغط الدم

blood vessel وعاء دموى

Buoyancy الطفو

Buoyant force قوة الطفو

8

Capacitance سعة المكثف

Capacitor المكثف

Coefficient of friction معامل األحتكاك

Charge شحنة

Circuit دائرة

cohesion قوة التماسك بين جزئيات السائل

components of a vector مكونات متجه

compression ضغط كباس أنضغاط

conduction توصيل

conductivity معامل الموصلية الحرارية

Conductor الموصل

Conservation of Energy حفظ الطاقة

conservation of momentum حفظ كمية الحركة

consumed مستهلك

Coulomb كولوم

cube مكعب

Current تيار

deceleration تباطؤ

Density الكثافة

Dendrites التشعبات العصبية

diffraction حيود الضوء

dimensions االبعاد

Dispersion تشتت تفرق

direction إتجاه

directly proportional يتناسب طرديا

displacement اإلزاحة

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

6

Week 13

Nerve Conduction

Ionic concentration and the resting potential

Response to weak stimuli The action potential

15

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 14 Daily Objectives SLO

Assignments amp Activities

Week 14

Wave properties of light

The Index of Refraction Reflection of Light Refraction of Light

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 15 Daily Objectives SLO

Assignments amp Activities

Week 15

Wave properties of light

Ionizing Radiation

The Interaction of Radiation with Matter

Chronic Radiation Exposure

16

- Lecturing

- Team work

- Exercises

- Self-learning

- Group Discussions

Week 16 Daily Objectives SLO

Assignments amp Activities

Week 16

Final Examination-Practical Lab

Week 17 Daily Objectives SLO

Assignments amp Activities

Week 17

Final Examination-

Comprehensive

11

12

13

14

15

16

22

23

24

31

32

Comprehensive Final Examination

Must be passed with a score of 60 or

better to pass course

7

Vocabulary

ENGLISH ARABIC

Acceleration تسارع ndashعجلة

Activity نشاط أشعاعى

Air pressure ضغط الهواء

Ampere أمبير

Analytical تحليلى

Analytical Method الطريق التحليلية

Angle زاوية

Angle of Deviation زاوية االنحراف

Angle of Incidence زواية السقوط

Angle of reflection زاوية االنعكاس

Angle of refraction زاوية األنكسار

Archimedess Principle مبدأ أرخميدس

Atmospheric pressure الضغط الجوي

Atom الذرة

Axoplasm جبلة المحوار

Axons محاور عصبية

Balanced Force القوة المتزنة

Bernoullis Principle law مبدأ )قانون( برنولي

Binding Energy طاقة الربط

Blood pressure ضغط الدم

blood vessel وعاء دموى

Buoyancy الطفو

Buoyant force قوة الطفو

8

Capacitance سعة المكثف

Capacitor المكثف

Coefficient of friction معامل األحتكاك

Charge شحنة

Circuit دائرة

cohesion قوة التماسك بين جزئيات السائل

components of a vector مكونات متجه

compression ضغط كباس أنضغاط

conduction توصيل

conductivity معامل الموصلية الحرارية

Conductor الموصل

Conservation of Energy حفظ الطاقة

conservation of momentum حفظ كمية الحركة

consumed مستهلك

Coulomb كولوم

cube مكعب

Current تيار

deceleration تباطؤ

Density الكثافة

Dendrites التشعبات العصبية

diffraction حيود الضوء

dimensions االبعاد

Dispersion تشتت تفرق

direction إتجاه

directly proportional يتناسب طرديا

displacement اإلزاحة

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

7

Vocabulary

ENGLISH ARABIC

Acceleration تسارع ndashعجلة

Activity نشاط أشعاعى

Air pressure ضغط الهواء

Ampere أمبير

Analytical تحليلى

Analytical Method الطريق التحليلية

Angle زاوية

Angle of Deviation زاوية االنحراف

Angle of Incidence زواية السقوط

Angle of reflection زاوية االنعكاس

Angle of refraction زاوية األنكسار

Archimedess Principle مبدأ أرخميدس

Atmospheric pressure الضغط الجوي

Atom الذرة

Axoplasm جبلة المحوار

Axons محاور عصبية

Balanced Force القوة المتزنة

Bernoullis Principle law مبدأ )قانون( برنولي

Binding Energy طاقة الربط

Blood pressure ضغط الدم

blood vessel وعاء دموى

Buoyancy الطفو

Buoyant force قوة الطفو

8

Capacitance سعة المكثف

Capacitor المكثف

Coefficient of friction معامل األحتكاك

Charge شحنة

Circuit دائرة

cohesion قوة التماسك بين جزئيات السائل

components of a vector مكونات متجه

compression ضغط كباس أنضغاط

conduction توصيل

conductivity معامل الموصلية الحرارية

Conductor الموصل

Conservation of Energy حفظ الطاقة

conservation of momentum حفظ كمية الحركة

consumed مستهلك

Coulomb كولوم

cube مكعب

Current تيار

deceleration تباطؤ

Density الكثافة

Dendrites التشعبات العصبية

diffraction حيود الضوء

dimensions االبعاد

Dispersion تشتت تفرق

direction إتجاه

directly proportional يتناسب طرديا

displacement اإلزاحة

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

8

Capacitance سعة المكثف

Capacitor المكثف

Coefficient of friction معامل األحتكاك

Charge شحنة

Circuit دائرة

cohesion قوة التماسك بين جزئيات السائل

components of a vector مكونات متجه

compression ضغط كباس أنضغاط

conduction توصيل

conductivity معامل الموصلية الحرارية

Conductor الموصل

Conservation of Energy حفظ الطاقة

conservation of momentum حفظ كمية الحركة

consumed مستهلك

Coulomb كولوم

cube مكعب

Current تيار

deceleration تباطؤ

Density الكثافة

Dendrites التشعبات العصبية

diffraction حيود الضوء

dimensions االبعاد

Dispersion تشتت تفرق

direction إتجاه

directly proportional يتناسب طرديا

displacement اإلزاحة

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

9

distance المسافة

Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك

عبره(

drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل

dynamics الديناميكا

efficiency كفاءة

effort arm ذراع القوة

Electric Charge شحنة كهربائية

Electric Circuit دائرة كهربائية

Electric Current التيار الكهربى

Electric Energy طاقة كهربائية

Electric field المجال الكهربي

Electrical Conductivity توصيل كهربائى

Electric Potential جهد كهربائى

Electrical Resistance مقاومة كهربائية

electricity الكهرباء

electrode قطب كهربائى

Electromagnetic Field مجال كهرومغناطيسي

electromagnetic induction الحث الكهرومغناطيسي

Electromotive Force القوة الدافعة الكهربية

Electron اإللكترون

Electron Diffraction حيود اإللكترون

Energy الطاقة

Energy Level مستويات الطاقة

Energy Transformations تحوالت الطاقة

Equation of continuity معادلة اإلستمرارية

Equilibrium إتزان

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

10

ev إلكترون فولت

Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل

Field مجال

Fluid المائع السائل

Fluid Dynamics ديناميكا الموائع

flow rate معدل السريان

Force قوة

force exerted القوة المبذولة

Frequency التردد

Friction اإلحتكاك

Friction forces قوة االحتكاك

fusion دمج -إنصهار

Geometric Method الرسم الهندسى او البيانىطريق

graph الرسم البيانى

Gravitational Force قوة الجاذبية

Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني

Gravity الجاذبية

Heat حرارة

Heat Energy الطاقة الحرارية

Heat Transfer انتقال الحرارة

Heavy Water الثقيلالماء

History of Physics تاريخ الفيزياء

Hookes Law قانون هوك

horizontal أفقى

Impedance المقاومه الكهربائيه

Ideal Gas Law قانون الغاز المثالي

inclined مائل

inertia القصور الذاتى

Infinity النهائية

intensity الشدة

interference التداخل

International System of Units نظام الوحدات الدولي

inversely proportional يتناسب عكسى

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

11

Ion أيون

ionic concentration التركيز األيوني

Ionizing التأين

Ionizing Radiation أشعة مؤينة

Joule )الجول )الوحدة الدولية لقياس الطاقة

Kelvin المطلقةكلفن درجة الحرارة

Kinetics علم الحركة

Kinetic Energy الطاقة الحركية

kinetic friction االحتكاك الحركي

Laminar flow تدفق المنار

laser الليزر

law of conservation of mechanical energy

قانون حفظ الطاقة الميكانيكية

laws of motion قوانين الحركة

leakage resistance مقاومة التسرب

Light year السنة الضوئية

Light الضوء

Liquid السائل

longitudinal wave الموجه الطوليه

Luminosity سطوع

Magnetic Field مجال مغناطيسي

Magnetic Flux تدفق مغناطيسي

Magnetic Moment عزم مغناطيسي

magnitude معيار amp قيمة

manometer ضغط الدم مقياس

mass الكتلة

matter مادة

mechanics ميكانيكا

mechanical energy الطاقة الميكانيكية

medium وسط

metal معدن

molecules جزئيات

Motion حركة

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

12

Movement حركة

net force محصلة القوة

neutron النيوترون

Nerve عصب

Nerve cells (neuron) خاليا عصبية

Nerve conduction التوصيل العصبى

newton نيوتن

Newtons first law قانون نيوتن األول

Newtons first law of motion قانون نيوتن األول للحركة

Newtons first law of motion (Inertia)

قانون نيوتن األول للحرآة قانون القصور الذاتي

newtons law of gravitation قانون نيوتن للجاذبية

newtons second law of motion قانون نيوتن الثاني للحركة

newtons third law of motion قانون نيوتن الثالث للحركة

normal force قوة عمودية

nuclear radiation اإلشعاع النووي

nucleus نواة

Ohm أوم

Ohms Law قانون

Optics البصريات

particle جسيم

pendulum بندول

photoelectric effect التأثير الكهروضوئي

photon فوتون

physics الفيزياء

pipeline خط انابيب

position موضع

Potential Difference فرق الجهد

Potential Energy طاقة الوضع

power القدرة

pressure الضغط

pressure in liquids في السوائل الضغط

Proton بروتون

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

13

Pulses ومضات

pull سحب

pulling force قوة السحب

quantity كمية

radiation اإلشعاع

radius نصف القطر

reaction رد الفعل

reflection أنعكاس

refraction األنكسار

refractive index معامل االنكسار الضوئي

relativity النسبية

resistance المقاومة

resistance force قوة مقاومة

resistivity المقاومة النوعية

resistor المقاوم

Rest Energy طاقة السكون

Resultant ناتج

Resultant Force القوة الناتجة

Scalar قياسى

Scalar quantities كميات قياسية

smooth أملس -ناعم

Smooth horizontal surface سطح أفقى املس

sound الصوت

space الفضاء

speed السرعة المطلقة

sphygmomanometer مقياس ضغط الدم

static electricity الكهريبة الساكنة

surface السطح

surface tension التوتر السطحى

tension توتر -شد

terminal velocity سرعة الوصول

Temperature درجة الحرارة

Thermal Physics الفيزياء الحرارية

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

14

thermometer مقياس الحرارة

Turbulence اضطراب

turbulent flow تدفق مضطرب

Transformers المحوالت

Transient Energy الطاقة الزائلة

Transverse Waves الموجات المستعرضة

Ultrasound فوق صوتى

ultraviolet ray االشعة فوق البنفسجية

unbalanced forces قوى غير متزنة

uniform motion حركة منتظمة

Vacuum الفراغ

valence electron إلكترونات التكافؤ

vector متجه

vectors متجهات

Vectors Geometry هندسة المتجهات

Velocity السرعة المتجهة

Viscosity اللزوجة

visual angle زاوية اإلبصار

volt فولت وحدة قياس فرق الجهد الكهربي

voltage الجهد الكهربي

voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي

Volume الحجم

water pipe انابيب مياه

water pressure ضغط الماء

watt W وحدة قياس القدرة الكهربية -واط

Wave موجة

Wave Function الدالة الموجية

wave length الطول الموجي

Wave Interference التداخل الموجى

Wave Motion الحركة الموجية

Wave Phenomena الظواهر الموجية

wave speed سرعة الموجة

Wave Superposition التراكب الموجى

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

15

weakcohesive

Weight الوزن

work الشغل

Work done الشغل المبذول

X-Ray )اشعة إكس ) السينية

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

16

Important unit conversions in physics course

Cm x 10-2 m

m m x 10-3m

μ m x 10-6m

Litter x 10-3m3

Gram x 10-3 kg

Symbol

L ndash length (m)

A ndash cross sectional area (m2)

r- radius (m)

R ndash resistance (Ω)

ρ (rho)ndash resistivity (Ω m) specific electrical

resistance

ρ=mv The density of a fluid(kgm3)

P ndash Pressure (Nm2 called the Pascal (Pa)

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

17

Ch 1 (11 Vector)

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

18

Part 1 Define scalar and vector quantity

Part 2 Adding vector

There are three methods to adding Vector

1- Graphical or called (Geometrical Method)

2- Pythagorean Theorem

3- Analytical Method or called Components Method

1- Graphical or called (Geometrical Method)

Add vectors A and B graphically by drawing them together in a head to

tail arrangement

Draw vector A first and then draw vector B such that its tail is on

the head of vector A

Then draw the sum or resultant vector by drawing a vector from the

tail of A to the head of B

Measure the magnitude and direction of the resultant vector

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

19

Example 1

A man walks at40 meters East and 30 meters north Find the magnitude

of resultant displacement and its vector angle Use Graphical Method

Answer

Given

A = 40 meters East B = 30 meters North

Resultant (R) = Angle θ =

So from this

Resultant (R) =50 amp Angle θ = 37

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

20

2- Pythagorean Theorem

The Pythagorean Theorem is a useful method for determining the result

of adding two (and only two) vectors and must be the angle between

this two vector equal =90

Example2

A man walks at 40 meters East and 30 meters North Find the magnitude

of resultant displacement and its vector angle Use Pythagorean

Theorem

Answer

_____________________________________________________

22 BABAR

)BA (1Tan

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

21

Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B

Solution

1- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A+B

Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o

We use the equation

Example

Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B

Solution

2- we find the total angle θ =θA-θB SO θ =120-60 = 60

2-We use the equation

So A-B

3-Analytical Method or called Components Method

First to calculate the components and magnitude of vector for example

the components of vector A are

Ax = A Cos θ and Ay = A sin θ

COSABBABA 222

COSABBABA 222

44106075275 22 COSxx

COSABBABA 222

COSABBABA 222

2466075275 22 COSxx

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

22

Example 1

Find the components of the vector A If A = 2 and the angle θ = 30o

Solution

Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173

Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1

Example 2

Given A = 3 and θ = 90o find Ax and Ay

Solution

Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0

Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3

Second To calculate the magnitude of vector for example magnitude vector A and

direction angle

We use the equation and

Example

If the components of a vector are defined by Ax =346 and Ay =2 find the

magnitude and direction angle of the vector A

Solution

1-We use the equation to find the magnitude vector

So the magnitude vector A=399

2- To find the direction angle we use the equation

30o So the direction angle θ=30o

22

yx AAA ) AA( xy1Tan

22

yx AAA

993)2()463( 22 A

) AA( xy1Tan

) 3462(1Tan

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

23

Third To calculate the resultant vector by component method

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

24

Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265

1- Calculate the Resultant magnitude by using component method

2- Calculate the Resultant angle direction

Answer

solution (1) We use the last equations So

By using equation so use the equation

solution (2) we use the equation

so

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

25

Part 3 Unit Vector Notation and product of vector

Unit Vector Notation

A unit vector is a vector that has a magnitude of one unit and can have any

direction

1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction

2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this

in two dimensions

3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z

direction

Notes

If AampB are two vectors where

A = axi + ayj + azkamp B = bxi + byj + bzk Then the

1- To findA+B and A B

A+B= (ax +bx)i + (ay +by)j + (az +bz)k

A B= (axbx)i + (ayby)j + (azbz)k

Example

Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B

Solution

1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k

So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k

2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k

So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j

_____________________________________________________

2-To find the magnitude of A+B and A B

Example 222 )()()( zzyyxx bababaBA

222 )()()( zzyyxx bababaBA

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

26

Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B

Solution

1- To find the magnitude for A+B

According the equation

So =1166

2- To find the magnitude for AB

According the equation

So =282

2-the magnitude of vector in Unit Vector Notation

If A is vectoring where A = axi + ayj + azk Then the

To find magnitude of vector Awe use the equation

Example

vector A = 3i +2j +3Kfind magnitude of vector A

Solution

According the last equation

So

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222 )()()( zzyyxx bababaBA

222 )33()42()53( BA

222

zyx aaaA

694323 222 A

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

27

Product of Vectors

There are two kinds of vector product

1 The first one is called scalar product or dot product because the result of

the product is a scalar quantity

2 The second is called vector product or cross product because the result is a

vector perpendicular to the plane of the two vectors

Example on the dot(scalar)and cross product

1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o

Find a) A B c) A x B

Solution

Θ=θBθA = 70o35o= 35o

So A B= A B COSθ = 4 x5 x COS 35o=1638

A x B= A B Sinθ = 4 x5 x Sin 35o=1147

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

28

Notes on the scalar product

If A amp B are two vectors where

A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk

Then their Scalar Product is defined as

AB = AxBx + AyBy + AzBz Where

amp

Example

Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B

Solution

According the last equation

So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

29

Summary low in the chapter

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

30

Quizzes

1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o

find a) A +b b) A - b c) A x B d) A B

2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B

3- A man walks at 20 meters East and 15 meters north Find the magnitude of

resultant displacement and its vector angle Use Graphical Method and

Pythagorean Theorem

4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o

Find a) A +b b) A - b c) A x B d) A B

5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB

6- Vector A has a magnitude of 5 units and direction angle ΘA = 30

find Ax and Ay

7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A

8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240

Calculate the Resultant magnitude by using component method

Calculate the Resultant angle direction

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

31

Choose the correct answer

Which of the following is a physical quantity that has a magnitude but no direction

A Vector B Resultant C Scalar D None

Which of the following is an example of a vector quantity

A Temperature B Velocity C Volume D Mass

Which of the following is a physical quantity that has a magnitude and direction

A Vector B Resultant C Scalar D None

Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17

The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N

Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|

A |B +A|= 5

B |B +A|=72

C |B +A|=1044

D |B +A|=86

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

32

A car travels 90 meters due north in 15 seconds Then the car turns around and

travels 40 meters due south What is the magnitude and direction of the cars

resultant displacement

A 40 meters South

B 50 meters South C 50 meters North

D 40 meters North

A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km

What is magnitude

A The direction that describes a quantity

B A numerical value C A unit of force

150N weight hanging DOWN from a rope Vector or scalar

A Scalar

B Vector

What type of quantity is produced by the dot product of two vectors

A scalar

B vector

Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB

A - 6

B - 8

C -2

D -3

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

33

Ch1 (12Newtons laws)

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

34

Facts about FORCE

Force unit is the NEWTON (N)

Its definition a push or a pull

What change the state of object is called ldquoforcerdquo

Means that we can control the magnitude of the

applied force and also its direction so force is a vector

quantity just like velocity and acceleration

Adding Forces

Forces are vectors (They have both magnitude and direction)

and so add as follows

1-Adding Forces In one dimension

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

35

2-Adding Forces In two dimensions

a) The angle between them is 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

2

2

2

1 FFF

NF 252015 22

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

36

B) The angle between them is or 90deg

Example

In this figure shown find the resultant (Net) force

Solution

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

37

Newtonrsquos First Law

An object at rest tends to stay at rest and an object in motion tends to

stay in motion with the same speed and in the same direction unless

an external force is acting on it

Or in other words

Everybody continuous in its state of rest or in uniform motion Unless

an external force is acting on it

Notes Newtonrsquos First Law is also called the Law of Inertia

So

Inertia is a term used to measure the ability of an object to

resist a change in its state of motion

An object with a lot of inertia takes a lot of force to start or

stop an object with a small amount of inertia requires a small

amount of force to start or stop

------------------------------------------------------------------------------------------------

Weight

Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)

mgW

Inability of an object to change its position by itself is called Inertia

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

38

Normal Forces Fn

Normal force this force acts in the direction perpendicular to the contact

surface and opposite the weight

Friction Forces Ff

Is opposing force caused by the interaction between two surfaces

Calculate the Friction Force and Normal Force

a) With out angle

Notes

If the surface is smooth the friction force Ff= 0

N

mg

F

N

w

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

39

Example 1

A man is pulling 20Kg suitcase with constant speed on a horizontal rough

floor show figure The pulling force F1 action is unknown Find The pulling

force F1 and normal force FN

Solution

From figure

F1= F2 = 20 N So the pulling force F1 action is 20 N

FN =m g where m= 20 and g=10 So FN = 20 x10=200N

Example 2

In this figure shown the object is at rest Find normal force FN

Solution

From figure

FN + F2 = F1 FN = F1F2 =2510=15 N

So the normal force FN =15 N

b) With angle

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

40

Example

An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude

of the pulling force F= 16N and its direct 30degabove the horizontal Find

a) Friction force b) The normal force FN

Solution

Given

m=5 Fp =16N θ=30deg and g=10

The pulling force F analysis in x and y direction show figure

a) Friction force

Ff = Fx = F cos θ=16 x cos30deg = 138 N

So Ff Friction force =138 N

b) The normal force FN

FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N

So FN normal force =42 N

---------------------------------------------------------------------

Newtonrsquos Second Law

ldquoForce equals mass times accelerationrdquo

F = ma

What does F = ma mean

Force is directly proportional to mass and acceleration

Notes

Newtonrsquos second law states that the net force on an object is

proportional to the mass and the acceleration that the object

undergoes

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

41

(a)Acceleration a measurement of how quickly an object is

changing speed a= Fm

Example

Calculate the force required to accelerate a 5Kg block along the floor at 30

ms2

Solution

Given m=5 and a= 30 ms2

According F = ma so F = 5x3=15 N

Net Force

The net force is the vector sum of all the forces acting on a body

321net FFFFF

aF m Example 1

The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg

Find

1 The net force

2 The acceleration of the block

Solution

1 we find the resultant (Net) force

According the equation

So

COSFFFFF 21

2

2

2

1 2

NCOSxxF 514301052105 22

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

42

2 The acceleration of the block (a)

a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2

Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of

50N Calculate the acceleration of the box if a 12 N frictional force acts upon it

Solution

Given m=10 Fa=50 and Ff=12

first we find the resultant (Net) force

So the acceleration of the box

------------------------------------------------------------------------------------------------------------

Newtonrsquos Third Law

ldquoFor every action there is an equal and opposite reactionrdquo

Coefficients of friction

Coefficient of friction is the ratio between friction force and normal force

Symbol is the Greek letter mu (μ)

μ= Ff FN

The coefficient of friction has no units

-----------------------------------------------------------------------------------------------------

Notes

Friction Force = Coefficient of friction Normal Force

Ffriction = Fnormal

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

43

Example1

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

Find the coefficient of friction between the bag and the floor

Solution

Given Fp=40 N m=20and g=10

From figure

Ff = Fp = 40 N So Ff action is 40 N

FN = m g where m= 20 and g=10 So FN = 20 x10=200N

So the coefficient of friction ( μ)

μ= Ff FN μ= 40 200 =02

-----------------------------------------------------------------------------------------------

Example 2

A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

1 What is the magnitude of the force of friction

2 What is the acceleration of the suit case

Solution

Given Fp=90 N m=30 g=10 ms2 and microk =01

1 Ff= microk FN where FN =m g=30times10=300 N

So Ff= 01times300=30 N so the magnitude of the force of friction = 30 N

2 The acceleration (a)

a= 119865119899119890119905

119898=

119865119901 minus119865119891

119898=

90minus30

30=

60

30= 2 1198981199042

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

44

Example3

A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N

inclined at 30deg above the horizontal and the coefficient of friction between the bag and

the floor is 01

a What is the magnitude of the force of friction

Given

m=20kg Fp =40N θ=30deg =01 and g=10

the pulling force F analysis in x and y direction show figure

Fx = F cos θ=40 x cos30deg = 346 N

Fy = FSin θ=40xsin 30deg= 20 N

FN=mg Fy=20x1020 = 180 N

Ff = FN Ff = 01 X 180 =18N

So the magnitude of the force of friction is 18N

bWhat is the acceleration of the suit case

119886 =119865119899119890119905

119898=

119865119909 minus 119865119891

119898=

346 minus 18

20=

166

20= 0083 1198981199042

so the acceleration of the suit case is 0083 mls2

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

45

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

46

Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m

2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force

and acceleration of the block

3 An object of mass m=3Kg is subject to a force F=9N Find

a) Wight of the object b) the acceleration of the object

4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the

resultant force

5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of

the force F= 16N and its direct 30degabove the horizontal Find

a) The normal force N b) The acceleration of the object

6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg

above the horizontal and the coefficient of friction between the bag and the floor is 01

What is the force of friction

What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of

350 N on the feet Find the force exerted by the chair on him

8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is

02What is the pulling force

9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is

experiencing 300 N force applied by the ground Find the force applied on him by the chair

10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of

friction is Fk = 60 N What is the coefficient of friction microk

11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and

the coefficient of friction microk =01

What is the magnitude of the force of friction

What is the acceleration of the suit case

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

47

Choose the correct answer 1 What type of forces do not change the motion of an object

a balanced forces

b unbalanced forces c static forces d accelerating forces

2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is

a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion

3 What unit do we use to measure force

a Newton b Meter c Pascal d Joule

4 When an unbalanced force acts on an object the force

a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object

5 What is the acceleration of gravity

a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2

6 An objects resistance to change in motion

b Motion c Inertia d Friction e Mass

7 is the measure of the force of gravity on an object

a mass b weight c density d equation

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

48

8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion

d Newtons fourth law of motion

9 The force of attraction between any two objects that have mass a Energy b Force c Gravity

d Speed

10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of

a Newtons First Law of Motion b Pascals Law

c Newtons Third Law of Motion d Archimedes Principle

11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)

a F=ma

b m=Fa c aF=m d m=aF

12 A force that resists motion created by objects rubbing together is a gravity

b friction c speed d force

13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface

The force of friction is Fk = 60 N What is the coefficient of friction microk

a) 05 b) 01

c) 03 d) 02

14 In the figure shown find the resultant (Net) force

a) 106 b) 2078

c) 145 d) 304

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

49

15For every action therersquos an equal and opposite reaction

a Newtons First Law

b Newtons Second Law c Newtons Third Law d Force

16The sum of all the forces acting on an object or system a net force b force

c normal force d drag force

17 an opposing force caused by the interaction between two surfaces

a inertia b mass c friction d force

18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass

19The force perpendicular to the surface that pushes up on the object of concern

a normal force

b force c drag force

d net force

20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N

d 0 N

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

50

Ch 2 work and energy

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

51

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

52

Notes on Work

Work = The Scalar Dot Product between Force F

and Displacement d

W = F d

The unit of work is a joule (J) and J = N middot m

Calculate work done on an object

1-Without angle

a) with apply force

The equation used to calculate the work (W) in this case it

W= F d

Example

How much work is done pulling with a 15 N force applied at

distance of 12 m

Solution

Given F=15 N amp d=12m

According the equation W= F d

So W=15x12=180 J

ntdisplacemeForceWork

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

53

b) Also with friction force

The equation used to calculate the work (W) in this case it

W= -Ff d -----------1

But Ffriction = Fnormal so you can write this equation (1)

W= -(Fnormal)d ---------2

But Fnormal= m g so you can write this equation(2)

W= -(mg)d ---------3

-------------------------------------------------------------------

Example

A horizontal force F pulls a 10 kg carton across the floor at

constant speed If the coefficient of sliding friction between the

carton and the floor is 030 how much work is done by F in

moving the carton by 5m

Solution

Given m=10 kg d=5m g=10 and μ=30 W=

The carton moves with constant speed Thus the carton is in

horizontal equilibrium

Fp = Ff = μk N = μk mg

Thus F = 03 x 10 x 10= 30 N

Therefore work done W = F d=30 x 5= 150 J

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

54

2-With angle

In this case the work done given by

Example

How much work is done pulling with a 15 N force applied at 20o over

a distance of 12 m

Solution

Given F=15 N θ=20oamp d=12m

According the equation W= F dCos θ

So W=15x12xCos 20o=1691 J

----------------------------------------------------------------------

Example

An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg

and he exerts a force of 120 times 102 N on the sled by pulling on the rope

a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) Suppose microk = 0200 How much work done on the sled by friction

c) Calculate the net work if θ = 30deg and he pulls the sled 50 m

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

55

Solution

Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10

a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m

b) calculate the work done on the sled by friction

c) Calculate the net work

J

mN

dFW

520

)30)(cos05)(10201(

cos

2

J

N

dFmgxN

dFxFW

kk

fffric

440

)5)(30sin10211050)(2000(

)sin(

)180cos(

2

J

WWWWW gNfricFnet

090

00440520

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

56

Kinetic Energy

Kinetic Energy is the energy of a particle due to its motion

KE = frac12 mv2

Where

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle

Also KE = frac12 mv2 so V2 =120784119948

119950 V=radic

120784119948

119950

Example 1 A 1500 kg car moves down the freeway at 30 ms Find the Kinetic Energy

Solution Given m=1500kg v=30ms

According the equation KE = frac12 mv2

So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

Example 2 A 10 kg mass has a kinetic energy of 25 joule What is the speed

Solution Given m=10 kg KE =20 joule v=

V=radic120784119948

119950=radic

120784119961120784120782

120783120782= radic

120786120782

120783120782= radic120786 = 2 ms

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

57

Work and Kinetic Energy

When work is done on a system and the only change in the

system is in its speed the work done by the net force equals

the change in kinetic energy of the system

So W = Kf - K0 ------------1

And also W =frac12 mvf2 frac12 m v0

2 ------------2

But W= -Ff d

So -Ff d=frac12 mvf2 frac12 m v0

2 ------------3

From equation (3) you can calculate the friction force

Example

A child of 40kg mass is running with speed 3ms on a

rough horizontal floor skids a distance 4 m till stopped

a) Find the force of friction

b) Find the coefficient of friction

Solution

Given m=40 kg v0=3ms vf=0 d= 4m and g=10

a) Calculate the force of friction

We apply the equation -Ff d=frac12 mvf2 frac12 m v0

2

But vf=0 so frac12 mvf2 =0

-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0

2

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

58

Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N

So the force of friction = 45 N

b) Calculate the coefficient of friction

According the equation in ch2 μ= Ff FN

Where Ff= 45 N and FN =mg=4010=400

So μ= Ff FN μ= 45 400 μ=01

---------------------------------------------------------------------------- Example

A 60-kg block initially at rest is pulled to the right along a horizontal

frictionless surface by a constant horizontal force of 12 N Find the speed of

the block after it has moved 30 m

Solution

GivenFp= 12 N m=6 kg v0=0 vf=

d= 3m and g=10

W =Fp d =12x 3 = 36J

Δk = w

frac12 mvf2 frac12 m v0

2 = w

But vo=0 so frac12 mv02 =0

frac12 mvf2 = W

frac12 x 6 x vf2 = 36 vf

sm 46312

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

59

Potential Energy

Potential Energy means the work done by gravity on the object

The formula for potential energy (U) due to gravity is U = mgh

PE = mass x height x gravity

The unit of Potential Energy is a joule (J)

----------------------------------------------------------------------------------------

Example

A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the

ground What is the potential energy of child

Solution

Given m=40 kg h= 60m and g=10

According the equation U = mgh

So U = 40 x 10x 60=24000 J

---------------------------------------------------------------------------------------------------

Conservation of Energy

bull Conservation of Mechanical Energy

MEi = MEf

initial mechanical energy = final mechanical energy

SO Ko + Uo = Kf + Uf----------1

SO Uo Uf = KfKo

So KfKo= -(Uf Uo )

K= U-----------------------------------2

So frac12 mvf2 frac12 m v0

2 = mg(hfho)---------------------------3

The equation (123) is very important

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

60

Example

At a construction site a 150 kg brick is dropped from rest and hit the ground

at a speed of 260 ms Assuming air resistance can be ignored calculate the

gravitational potential energy of the brick before it was dropped

Solution

Given m=150 kg v0=0 vf=26 Uf=0 Uo=

According Ko + Uo = Kf + Uf

But vo=0 so Ko =frac12 mv02 =0 and Uf=0

So Uo = Kf Uo=mgho = frac12 mvf2

Uo= frac12 x (15x 26)2= 507 J

-----------------------------------------------------------------------------------------------

Example

A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m

above her lowest position Find her speed at the lowest position

Solution

Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10

According the equation frac12 mvf2 frac12 m v0

2 = mg(hfho)

But vo=0 so frac12 mv02 =0

frac12 mvf2 = mg(03) frac12 mvf

2 = 3mg frac12 vf2 = 3g

g=10 frac12 vf2 = 30 vf

2 = 60

vf sm 7760

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

61

Power Power is is the rate of doing work It is the amount of energy consumed per

unit time

P =119882

119905=

119865119889

119905= 119865V where V=dt

Units of Power

Where the unit of work(W) is joule and unit of time(t) is second So The

unit of power is a Watt

where 1 watt = 1 joule second

--------------------------------------------------------------------------------------

Example

A 100 N force is applied to an object in order to lift it a distance of 20 m

within 60 s Find the power

Solution

Given F=100 N d=20 m t=60 s

According the equation P =119882

119905=

119865119889

119905=

100 119857 20

60=

3333 waat

62

Example

A woman of 50 Kg mass climbs a mountain 4000 m high

a) Find the work she did against gravitational forces

b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to

energy with efficiency rate of 25 How much fat she consumed in the climb

Solution

Given m=50 kg h=4000 m and g= 10 ms2

a) Calculate the work she did against gravitational forces

W= F d where in this case F= m g and d=h

So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J

b) Calculate the fat consumed in the climb

According the equation 119862 = 119882119905

where W=2 x 106 J

And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106

So 119862 = 119882119905

=2 x 106

925 x 106= 0216 kg

63

Example

A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance

of 05m 1000 times

a) Find the work he did against gravitational forces

b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the

man converts energy at 20 efficiency rate How much fat will he

consume in the exercise

Solution

Given mass for man m1=70 kg and he is lifting mass m2 =10

So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2

a) Calculate the work he did against gravitational forces

W= F d where in this case F= m g

So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces

W=400x1000=400000 J=4 x 105 J

b) Calculate the fat consumed in the exercise

According the equation 119862 = 119882119905

where W=4 x 105 J

And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105

So 119862 = 119882119905

=4 x 105

76 x 105 = 00526 kg

64

Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground

2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant

speed The force of friction between the box and the surface is 20 N

Find the work it did against the force of friction

3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high

Find the speed of the ball 20 m above the base of the tower

4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal

road Find the energy loss

5 A boy of 50 Kg mass climbrsquos a wall 500 m high

a) Find the work he did against gravitational forces

b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency

rate of 25 How much fat he consumed in the climb

6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal

road Find the force of friction

7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after

smashing 15 meter of the front of the car Find the reactive force acting on the car body during the

crash

8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times

a) Find the work he did against gravitational forces

b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with

efficiency rate of 25 How much fat he consumed in the exercise

9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m

till stopped Find the force of friction

10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at

the top of the tower

11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after

smashing 15 meter of the front of the car

a) Find the kinetic energy of the car

b) Find the reactive force acting on the car body during the crash

12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the

acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower

65

Choose the correct answer

Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy

Work done = Force x _______ A distance

B acceleration

C velocity

D speed

1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2

The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second

A watt per second B joule C kilojoule D joule per second

A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)

A 50 J

B 60 J

C 500 J

D 600 J

A B C D

66

A 1 kg mass has a kinetic energy of 1 joule when its speed is

A 045 ms

B ms

C 14 ms

D 44 ms

Name the physical quantity which is equal to the product of force and

distance

A Work

B energy

C power

D acceleration

An object of mass 1 kg has potential energy of 1 joule relative to the

ground when it is at a height of _______

A 010 m

B 1 m

C 98 m

D 32 m

What is kinetic energy

A When an object is in motion

B When an object is not in motion

C all of the above

D none of the above

It takes 20 N of force to move a box a distance of 10 m How much work is

done on the box A 200 J B 200J

C 2 J D 200 N

Two factors that determine work are

A amount of the force and effort used B amount of the force and type of force

C mass and distance D amount of force and distance moved

67

What is energy

A It is measured in watts B It is power

C It is the ability to do work D It is fluid motion

What is work A The product of force and displacement

B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position

D All of these are true

The law of conservation of energy states

A Energy cannot be created

B Energy cannot be destroyed C Energy can only be transferred

D All of these

68

Ch 3 THE MECHANICS OF NON-VISCOUS

FLUIDS

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

63

Example

A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance

of 05m 1000 times

a) Find the work he did against gravitational forces

b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the

man converts energy at 20 efficiency rate How much fat will he

consume in the exercise

Solution

Given mass for man m1=70 kg and he is lifting mass m2 =10

So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2

a) Calculate the work he did against gravitational forces

W= F d where in this case F= m g

So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces

W=400x1000=400000 J=4 x 105 J

b) Calculate the fat consumed in the exercise

According the equation 119862 = 119882119905

where W=4 x 105 J

And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105

So 119862 = 119882119905

=4 x 105

76 x 105 = 00526 kg

64

Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground

2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant

speed The force of friction between the box and the surface is 20 N

Find the work it did against the force of friction

3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high

Find the speed of the ball 20 m above the base of the tower

4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal

road Find the energy loss

5 A boy of 50 Kg mass climbrsquos a wall 500 m high

a) Find the work he did against gravitational forces

b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency

rate of 25 How much fat he consumed in the climb

6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal

road Find the force of friction

7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after

smashing 15 meter of the front of the car Find the reactive force acting on the car body during the

crash

8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times

a) Find the work he did against gravitational forces

b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with

efficiency rate of 25 How much fat he consumed in the exercise

9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m

till stopped Find the force of friction

10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at

the top of the tower

11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after

smashing 15 meter of the front of the car

a) Find the kinetic energy of the car

b) Find the reactive force acting on the car body during the crash

12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the

acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower

65

Choose the correct answer

Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy

Work done = Force x _______ A distance

B acceleration

C velocity

D speed

1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2

The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second

A watt per second B joule C kilojoule D joule per second

A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)

A 50 J

B 60 J

C 500 J

D 600 J

A B C D

66

A 1 kg mass has a kinetic energy of 1 joule when its speed is

A 045 ms

B ms

C 14 ms

D 44 ms

Name the physical quantity which is equal to the product of force and

distance

A Work

B energy

C power

D acceleration

An object of mass 1 kg has potential energy of 1 joule relative to the

ground when it is at a height of _______

A 010 m

B 1 m

C 98 m

D 32 m

What is kinetic energy

A When an object is in motion

B When an object is not in motion

C all of the above

D none of the above

It takes 20 N of force to move a box a distance of 10 m How much work is

done on the box A 200 J B 200J

C 2 J D 200 N

Two factors that determine work are

A amount of the force and effort used B amount of the force and type of force

C mass and distance D amount of force and distance moved

67

What is energy

A It is measured in watts B It is power

C It is the ability to do work D It is fluid motion

What is work A The product of force and displacement

B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position

D All of these are true

The law of conservation of energy states

A Energy cannot be created

B Energy cannot be destroyed C Energy can only be transferred

D All of these

68

Ch 3 THE MECHANICS OF NON-VISCOUS

FLUIDS

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

64

Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground

2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant

speed The force of friction between the box and the surface is 20 N

Find the work it did against the force of friction

3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high

Find the speed of the ball 20 m above the base of the tower

4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal

road Find the energy loss

5 A boy of 50 Kg mass climbrsquos a wall 500 m high

a) Find the work he did against gravitational forces

b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency

rate of 25 How much fat he consumed in the climb

6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal

road Find the force of friction

7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after

smashing 15 meter of the front of the car Find the reactive force acting on the car body during the

crash

8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times

a) Find the work he did against gravitational forces

b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with

efficiency rate of 25 How much fat he consumed in the exercise

9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m

till stopped Find the force of friction

10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at

the top of the tower

11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after

smashing 15 meter of the front of the car

a) Find the kinetic energy of the car

b) Find the reactive force acting on the car body during the crash

12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the

acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower

65

Choose the correct answer

Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy

Work done = Force x _______ A distance

B acceleration

C velocity

D speed

1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2

The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second

A watt per second B joule C kilojoule D joule per second

A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)

A 50 J

B 60 J

C 500 J

D 600 J

A B C D

66

A 1 kg mass has a kinetic energy of 1 joule when its speed is

A 045 ms

B ms

C 14 ms

D 44 ms

Name the physical quantity which is equal to the product of force and

distance

A Work

B energy

C power

D acceleration

An object of mass 1 kg has potential energy of 1 joule relative to the

ground when it is at a height of _______

A 010 m

B 1 m

C 98 m

D 32 m

What is kinetic energy

A When an object is in motion

B When an object is not in motion

C all of the above

D none of the above

It takes 20 N of force to move a box a distance of 10 m How much work is

done on the box A 200 J B 200J

C 2 J D 200 N

Two factors that determine work are

A amount of the force and effort used B amount of the force and type of force

C mass and distance D amount of force and distance moved

67

What is energy

A It is measured in watts B It is power

C It is the ability to do work D It is fluid motion

What is work A The product of force and displacement

B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position

D All of these are true

The law of conservation of energy states

A Energy cannot be created

B Energy cannot be destroyed C Energy can only be transferred

D All of these

68

Ch 3 THE MECHANICS OF NON-VISCOUS

FLUIDS

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

65

Choose the correct answer

Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy

Work done = Force x _______ A distance

B acceleration

C velocity

D speed

1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2

The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second

A watt per second B joule C kilojoule D joule per second

A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)

A 50 J

B 60 J

C 500 J

D 600 J

A B C D

66

A 1 kg mass has a kinetic energy of 1 joule when its speed is

A 045 ms

B ms

C 14 ms

D 44 ms

Name the physical quantity which is equal to the product of force and

distance

A Work

B energy

C power

D acceleration

An object of mass 1 kg has potential energy of 1 joule relative to the

ground when it is at a height of _______

A 010 m

B 1 m

C 98 m

D 32 m

What is kinetic energy

A When an object is in motion

B When an object is not in motion

C all of the above

D none of the above

It takes 20 N of force to move a box a distance of 10 m How much work is

done on the box A 200 J B 200J

C 2 J D 200 N

Two factors that determine work are

A amount of the force and effort used B amount of the force and type of force

C mass and distance D amount of force and distance moved

67

What is energy

A It is measured in watts B It is power

C It is the ability to do work D It is fluid motion

What is work A The product of force and displacement

B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position

D All of these are true

The law of conservation of energy states

A Energy cannot be created

B Energy cannot be destroyed C Energy can only be transferred

D All of these

68

Ch 3 THE MECHANICS OF NON-VISCOUS

FLUIDS

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

66

A 1 kg mass has a kinetic energy of 1 joule when its speed is

A 045 ms

B ms

C 14 ms

D 44 ms

Name the physical quantity which is equal to the product of force and

distance

A Work

B energy

C power

D acceleration

An object of mass 1 kg has potential energy of 1 joule relative to the

ground when it is at a height of _______

A 010 m

B 1 m

C 98 m

D 32 m

What is kinetic energy

A When an object is in motion

B When an object is not in motion

C all of the above

D none of the above

It takes 20 N of force to move a box a distance of 10 m How much work is

done on the box A 200 J B 200J

C 2 J D 200 N

Two factors that determine work are

A amount of the force and effort used B amount of the force and type of force

C mass and distance D amount of force and distance moved

67

What is energy

A It is measured in watts B It is power

C It is the ability to do work D It is fluid motion

What is work A The product of force and displacement

B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position

D All of these are true

The law of conservation of energy states

A Energy cannot be created

B Energy cannot be destroyed C Energy can only be transferred

D All of these

68

Ch 3 THE MECHANICS OF NON-VISCOUS

FLUIDS

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

67

What is energy

A It is measured in watts B It is power

C It is the ability to do work D It is fluid motion

What is work A The product of force and displacement

B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position

D All of these are true

The law of conservation of energy states

A Energy cannot be created

B Energy cannot be destroyed C Energy can only be transferred

D All of these

68

Ch 3 THE MECHANICS OF NON-VISCOUS

FLUIDS

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

68

Ch 3 THE MECHANICS OF NON-VISCOUS

FLUIDS

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

69

----------------------------------------------------------------------------

What is the Fluids

A fluid is a collection of molecules that are randomly arranged

and held together by weakcohesive forces and by forces exerted

by the Walls of a container

Both liquids and gases fluids

--------------------------------------------------------------

Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume

ρ=mv (uniform density)

bullDensity is a scalar the SI unit is kgm3

2-Pressure

P=FA (Pressure of uniform force on flat area)

bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

70

---------------------------------------------------------------

if there is an incompressible fluid completely fills a channel such as a pipe or an artery

Then if more fluid enters one end of the channel So an equal amount must leave the other

end This principle is called

The Equation of Continuity

The Equation of Continuity (STREAMLINE FLOW)

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

71

The flow rate (Q)

119876 is The flow rate which is the volume ΔV of the fluid flowing past a

point in a channel per unit time Δt

The SI unit of the flow rate 119876 is the 119950 3 119956

Example

If the volume of water flows flowing past a point in pipeline in 3

minutes is 5 litters what is the flow rat

Answer

Given

ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s

So according the last equation

Q = 119881

119905=

5x10minus3

180= 27x10minus5 1198983119852

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

72

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

73

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

74

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

75

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

76

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

77

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

78

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

79

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

80

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

81

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

82

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

83

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

84

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

85

Exercise 1 Questions and Answers

What are fluids A Substances that flow B Liquids and gases

C Aampb D Non of the above

Bernoullis principle states that for streamline motion of an incompressible

non-viscous fluid

A pressure at any part + kinetic energy per unit volume = constant

B kinetic energy per unit volume + potential energy per unit volume = constant

C pressure at any part + potential energy per unit volume = constant

D pressure at any part + kinetic energy per unit volume + potential energy per

unit volume = constant

If layers of fluid has frictional force between them then it is known as

A viscous

B non-viscous

C incompressible

D both a and b

If every particle of fluid has irregular flow then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

if every particle of fluid follow same path then flow is said to be

A laminar flow

B turbulent flow

C fluid flow

D both a and b

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

86

Which of the following is a fluid

A helium B ice

C iron D gold

Which of the following is NOT a fluid A carbon dioxide

B hydrogen C seawater

D wood

Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable

2 What is the fluid

3 What is the flow rate

4 Write the equation of continuity

5 Write the Bernoullis equation

6 The brain of a man is 05 m above his heart level The blood density ρ =10595

Kgm3What is the blood pressure difference between the brain and the heart

7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the

other end of radius 01 Cm find the velocity of blood out

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

87

Ch4 Direct currents

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

88

Electric current The electric current in a wire is the rate at which the charge moves in the wire

Definition of the current

The SI Current unit is the ampere (A)

t

QI

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

89

Ohmrsquos Law

For many conductors current depends on

Voltage - more voltage more current

Current is proportional to voltage

Resistance - more resistance less current

Current is inversely proportional to resistance

Example 3

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

90

Example 4

What is the resistance of the heating element in a car lock deicer that

contains a 15-V battery supplying a current of 05 A to the circuit

Resistance (R)

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

91

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

92

According to Ohms law Resistance is equal to to voltage divided by

A potential difference B conduction

C time D current

What is a circuit

A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source

C A pathway that electricity flows in It has a load wire and a power source

D A pathway that electricity flows in It has a load and wire

What is an Electric Current

A A An Electric Field B B An Ampere

C C The flow of electric charge

What is Ohms Law

A I=VR

B R=VI

C Power= Voltage times Current D AampB

A closed path that electric current follows A Voltage

B Current C Resistance

D Circuit

This is related to the force that causes electric charges to flow

A Voltage B Current

C Resistance D Circuit

What charge does an electron have

A negative (-) B positive (+)

C neutral or no charge (0)

Resistance is affected by a materialrsquos

A temperature B thickness

C length D all of these

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

93

The number of electrons flowing is called

A voltage B power C current D resistance

When the circuit is______ current does not flow

A resistors B heat C closed D open

Electrons leave the ______ of a battery and enter the ______ of the battery

A Positive terminal positive terminal

B Negative terminal negative terminal C Negative terminal positive terminal

D Positive Terminal Negative Terminal

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

94

Ch5 Nerve Conduction

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

95

Nerve Conduction

What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is

a measurement of the speed of conduction of an electrical impulse through a nerve

NCS can determine nerve damage and destruction

A nerve conduction study (NCS) is a medical diagnostic test commonly used to

evaluate the function especially the ability of electrical conduction of the motor and

sensory nerves of the human body

The structure of the nerve cells (neuron)

96

Nerve electric properties

97

98

99

100

101

102

103

104

96

Nerve electric properties

97

98

99

100

101

102

103

104

97

98

99

100

101

102

103

104

98

99

100

101

102

103

104

99

100

101

102

103

104

100

101

102

103

104

101

102

103

104

102

103

104

103

104

104